\(n_{C_2H_4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ PTHH:C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\\ Mol:0,2\rightarrow0,6\rightarrow0,4\rightarrow0,4\\ \rightarrow\left\{{}\begin{matrix}V_{kk}=0,6.5.22,4=67,2\left(l\right)\\V_{CO_2}=0,4.44=17,6\left(g\right)\\m_{H_2O}=0,4.18=7,2\left(g\right)\end{matrix}\right.\)